Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take the curve y equals x, times x minus one, times x minus two. It cuts the x axis at zero, one and two, making two little loops. Integrate it from zero to two, and you get exactly zero. So the area enclosed is zero, right? But the two loops are right there! Something is wrong, isn't it? Today, Set three of our mixed problems. Questions like these come up often in JEE Main, so let's begin.

0:33Tool 1: the integral is signed, the area is not

First tool. The long S is the integral sign, a and b are its limits, and d x says we integrate in x. This integral is the signed area between the curve and the x axis. Okay? Now the picture from the course. The dark blue curve is sine x, from zero to two pi; pi is the Greek letter. The pink part above the axis is marked plus two, and the grey part below, minus two. And the first line inside the picture says the integral of sine x, d x, from zero to two pi, is zero. But there is clearly some area there. So why zero? Because the part below the axis counts as negative, exactly! Plus two and minus two cancel. Right? But an area is never negative. So the area is the integral of mod f of x; the two straight bars are the modulus. For sine x that gives four, the second line inside the picture. Clear? So, split where f changes sign, and flip the sign of every negative piece. Look, this is the classic trap: integrate straight across the axis, and the pieces cancel, sometimes all the way to zero. Got it?

1:45Tool 2: fold the integral from 0 to 2a at the midpoint

Second tool. The integral from zero to two a of f of x equals the integral from zero to a of f of x, plus f of two a minus x; the square brackets just group the sum. We fold the interval at its midpoint, a: each x meets its partner, two a minus x. With me? If f of two a minus x equals f of x, the two halves match, and we get two times the integral from zero to a. Okay? And what if f of two a minus x equals minus f of x? What happens to the integral then? Yes, it is zero! Each value cancels its partner. Keep this one in mind. It is the secret behind our puzzle. Right?

2:27Tool 3: an x in front, limits 0 to pi

Third tool: the King property. The integral from a to b of f of x equals the integral from a to b of f of a plus b minus x, with a and b the limits of this same integral. Clear? Now take the limits zero and pi. Then a plus b minus x is just pi minus x. So tell me: what is sine of pi minus x? Sine x itself, correct! The arrow on screen means replace x by pi minus x. So f of sine x stays the same; only the x in front changes, to pi minus x. With me? Now add the two forms. x plus pi minus x is just pi, a constant, so it comes outside; then divide by two. So the integral of x times f of sine x, from zero to pi, is pi over two times the integral of f of sine x. The conditions: limits zero and pi, and the rest a function of sine x only. Okay?

3:23Example 1: area of y = x(x-1)(x-2) is 1/2, not 0

Now, Example one: our puzzle. Find the total area enclosed between the curve y equals x, times x minus one, times x minus two, and the x axis. Pause the video and try it first. Done? Okay, let's check. Call the curve f of x; it meets the axis at zero, one and two. f is greater than zero on zero to one, and less than zero on one to two; the round brackets mean open intervals. Right? So, one loop above the axis and one below. Now, why is the integral from zero to two exactly zero? Which tool explains it? Tool two, exactly! Here two a is two, so a is one. Replace x by two minus x: two minus x, times one minus x, times minus x, is minus f of x. The double arrow means so. The lower loop mirrors the upper one, and the signed integral cancels. Puzzle solved: zero is the signed area, not the area! Got it? So the area is the integral of the modulus of f of x. The two loops are equal, so take two times the loop from zero to one, where f is positive. Okay? Expand: x cubed, minus three x squared, plus two x. Integrate: x to the power four over four, minus x cubed, plus x squared. The square brackets mean the value at one minus the value at zero: one quarter, minus one, plus one, which is one quarter. Clear? So the area is two times one quarter, which is one half; the dot means times. Integrate straight from zero to two, and you would write zero, the trap from our puzzle. Very good!

5:08Example 2: x/(cos squared x + 4 sin squared x) from 0 to pi

Now a harder one, JEE Main level. Evaluate the integral from zero to pi of x, over cos squared x plus four sine squared x, d x. Pause the video and try it first. Done? Okay, let's check. Tool three needs the rest to be a function of sine x only. So, can we write the bottom using sine x alone? Yes, we can! Cos squared x is one minus sine squared x, so the bottom is one plus three sine squared x. Only sine x, so tool three applies. Right? By tool three, our integral, capital I, is pi over two, times the integral from zero to pi of one over the same bottom. This integrand does not change when x becomes pi minus x, so by tool two, with a equal to pi over two, it is twice the integral from zero to pi over two. Pi over two, times two, gives pi. With me? Now, why did we halve? Why not put t equals tan x straight away, from zero to pi? Tan zero and tan pi are both zero, so the answer is zero, right? No, no, careful here! The square brackets mean the closed interval from zero to pi, and tan x is not even defined at pi over two, right in its middle. So we halve first, and only then substitute. Okay? Now divide the top and the bottom by cos squared x. The top becomes sec squared x, and the bottom becomes one plus four tan squared x. Got it? Put t equals tan x, so d t equals sec squared x, d x. New limits: zero stays zero, and as x goes to pi over two, t goes to infinity, the sideways eight. The standard tan inverse result gives one half, tan inverse of two t; tan to the power minus one means tan inverse. Clear? Tan inverse of infinity is pi over two, and tan inverse of zero is zero. So capital I equals pi, times one half, times pi over two, which is pi squared over four. Not zero, as the trap said. That's it!

7:17Quick revision: area is mod f, fold at a, halve before tan x

So, quick revision. Area: integrate the modulus, splitting where f changes sign. From zero to two a: twice the integral from zero to a if f of two a minus x equals f of x, and zero if it equals minus f of x. With an x in front, from zero to pi, the x goes and pi over two comes out; and halve before putting t equals tan x. Now go and try the practice problems on this page. All the best!